Integrand size = 33, antiderivative size = 171 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {4 a^3 (9 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (3 A+5 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a^3 (21 A+20 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (9 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
-4/5*a^3*(9*A+5*B)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^3*(3*A+5* B)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+4/15*a^3*(21*A+20*B)*sin(d*x+c )/d/cos(d*x+c)^(1/2)+2/5*a*A*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(5 /2)+2/15*(9*A+5*B)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.22 (sec) , antiderivative size = 890, normalized size of antiderivative = 5.20 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(7/2) ,x]
Output:
Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/40*((-3 6*A - 25*B + 5*B*Cos[2*c])*Csc[c]*Sec[c])/d + (A*Sec[c]*Sec[c + d*x]^3*Sin [d*x])/(20*d) + (Sec[c]*Sec[c + d*x]^2*(3*A*Sin[c] + 15*A*Sin[d*x] + 5*B*S in[d*x]))/(60*d) + (Sec[c]*Sec[c + d*x]*(15*A*Sin[c] + 5*B*Sin[c] + 54*A*S in[d*x] + 45*B*Sin[d*x]))/(60*d)) - (A*(a + a*Cos[c + d*x])^3*Csc[c]*Hyper geometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d* x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt [-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(2*d*Sqrt[1 + Cot[c]^2]) - (5*B*(a + a*Cos[c + d*x])^3 *Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]* Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[C ot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt [1 + Sin[d*x - ArcTan[Cot[c]]]])/(6*d*Sqrt[1 + Cot[c]^2]) + (9*A*(a + a*Co s[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4} , {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(S qrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqr t[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Co s[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Co s[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(20*d) + (B*(a + a...
Time = 1.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3454, 27, 3042, 3454, 3042, 3447, 3042, 3500, 27, 3042, 3227, 3042, 3119, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {2}{5} \int \frac {(\cos (c+d x) a+a)^2 (a (9 A+5 B)-a (A-5 B) \cos (c+d x))}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(\cos (c+d x) a+a)^2 (a (9 A+5 B)-a (A-5 B) \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (9 A+5 B)-a (A-5 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {(\cos (c+d x) a+a) \left (a^2 (21 A+20 B)-a^2 (6 A-5 B) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (21 A+20 B)-a^2 (6 A-5 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {-\left ((6 A-5 B) \cos ^2(c+d x) a^3\right )+(21 A+20 B) a^3+\left (a^3 (21 A+20 B)-a^3 (6 A-5 B)\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {-\left ((6 A-5 B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+(21 A+20 B) a^3+\left (a^3 (21 A+20 B)-a^3 (6 A-5 B)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (2 \int \frac {5 a^3 (3 A+5 B)-3 a^3 (9 A+5 B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a^3 (21 A+20 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\int \frac {5 a^3 (3 A+5 B)-3 a^3 (9 A+5 B) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^3 (21 A+20 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (\int \frac {5 a^3 (3 A+5 B)-3 a^3 (9 A+5 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (21 A+20 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+5 B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (9 A+5 B) \int \sqrt {\cos (c+d x)}dx+\frac {2 a^3 (21 A+20 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+5 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (9 A+5 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^3 (21 A+20 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \left (5 a^3 (3 A+5 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^3 (9 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^3 (21 A+20 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (9 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \left (\frac {10 a^3 (3 A+5 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^3 (9 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^3 (21 A+20 B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )\right )+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(7/2),x]
Output:
(2*a*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2 *(9*A + 5*B)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2 )) + (2*((-6*a^3*(9*A + 5*B)*EllipticE[(c + d*x)/2, 2])/d + (10*a^3*(3*A + 5*B)*EllipticF[(c + d*x)/2, 2])/d + (2*a^3*(21*A + 20*B)*Sin[c + d*x])/(d *Sqrt[Cos[c + d*x]])))/3)/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(915\) vs. \(2(158)=316\).
Time = 9.07 (sec) , antiderivative size = 916, normalized size of antiderivative = 5.36
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(916\) |
| parts | \(\text {Expression too large to display}\) | \(946\) |
Input:
int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x,method=_RETURNV ERBOSE)
Output:
-4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^3*(216*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*A*(2*sin( 1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-108*A*(sin(1/2*d*x+1/2*c)^2)^(1/2) *EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*si n(1/2*d*x+1/2*c)^4+180*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-100*B*(si n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2 *d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-60*B*(2*sin(1/2*d*x+1/2*c)^2-1 )^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)) *sin(1/2*d*x+1/2*c)^4-246*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+60*A*( 2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos (1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+108*A*(sin(1/2*d*x+1/2*c)^2) ^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 /2)*sin(1/2*d*x+1/2*c)^2-190*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+100 *B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*s in(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+60*B*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^ (1/2))*sin(1/2*d*x+1/2*c)^2+72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-1 5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip...
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.42 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (9 \, {\left (6 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 3 \, A a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorith m="fricas")
Output:
-2/15*(5*I*sqrt(2)*(3*A + 5*B)*a^3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 5*B)*a^3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt( 2)*(9*A + 5*B)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(9*A + 5*B)*a^3*co s(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c ) - I*sin(d*x + c))) - (9*(6*A + 5*B)*a^3*cos(d*x + c)^2 + 5*(3*A + B)*a^3 *cos(d*x + c) + 3*A*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^ 3)
Timed out. \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c))/cos(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorith m="maxima")
Output:
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)
\[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(7/2), x)
Time = 43.60 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.68 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2\,\left (B\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,B\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {2\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^3)/cos(c + d*x)^(7/2),x)
Output:
(2*(B*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*B*a^3*ellipticF(c/2 + (d*x)/2, 2 )))/d + (2*A*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*A*a^3*sin(c + d*x)*hy pergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*A*a^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*A*a^3*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2 )*(sin(c + d*x)^2)^(1/2)) + (6*B*a^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3 /4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*B* a^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))
\[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=a^{3} \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) b +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) b \right ) \] Input:
int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x)
Output:
a**3*(int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a + 3*int(sqrt(cos(c + d*x))/ cos(c + d*x),x)*b + int(sqrt(cos(c + d*x))/cos(c + d*x)**4,x)*a + 3*int(sq rt(cos(c + d*x))/cos(c + d*x)**3,x)*a + int(sqrt(cos(c + d*x))/cos(c + d*x )**3,x)*b + 3*int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*a + 3*int(sqrt(cos (c + d*x))/cos(c + d*x)**2,x)*b + int(sqrt(cos(c + d*x)),x)*b)